The simple way of calculating a perimeter is to add all of the sides.
For a rectangle this would equal 2 x length + 2 x width
P = 2(6x + 3) + 2(-2x - 5)
= 12x + 6 - 4x - 10
= 8x - 4
At a higher level both the length and width must be greater than zero (= zero is a trivial rectangle)
6x + 3 > 0
6x > -3
x > -0.5
-2x - 5 > 0
2x + 5 < 0 (multiplying by -1 reverses the inequality)
2x < -5
x < -2.5
This rectangle cannot exist as x cannot be < -2.5 and > -0.5 at the same time!
Answer:
10
Explanation:
The distance between two points of coordinates (x1, y1) and (x2, y2) can be calculated as:
![\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
So, the distance between (-2, 5) and (8, 5) is equal to:
![\begin{gathered} \sqrt[]{(8-(-2))^2+(5-5)^2} \\ \sqrt[]{(8+2)^2+(0_{})^2} \\ \sqrt[]{(10)^2+0} \\ \sqrt[]{100} \\ 10 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csqrt%5B%5D%7B%288-%28-2%29%29%5E2%2B%285-5%29%5E2%7D%20%5C%5C%20%5Csqrt%5B%5D%7B%288%2B2%29%5E2%2B%280_%7B%7D%29%5E2%7D%20%5C%5C%20%5Csqrt%5B%5D%7B%2810%29%5E2%2B0%7D%20%5C%5C%20%5Csqrt%5B%5D%7B100%7D%20%5C%5C%2010%20%5Cend%7Bgathered%7D)
Then, the distance between (-2, 5) and (8, 5) is 10.
There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:
Consider an athlete running a 40-m dash. The position of the athlete is given by
where d is the position in meters and t is the time elapsed, measured in seconds.
Compute the average velocity of the runner over the intervals:
(a) [1.95, 2.05]
(b) [1.995, 2.005]
(c) [1.9995, 2.0005]
(d) [2, 2.00001]
Answer
(a) 6.00041667m/s
(b) 6.00000417 m/s
(c) 6.00000004 m/s
(d) 6.00001 m/s
The instantaneous velocity of the athlete at t=2s is 6m/s
Step by step Explanation:
In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:
so let's take the first interval:
(a) [1.95, 2.05]
we get that:
so:
(b) [1.995, 2.005]
we get that:
so:
(c) [1.9995, 2.0005]
we get that:
so:
(d) [2, 2.00001]
we get that:
so:
Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s
Answer:
I am not sure but for me the answer is 1
Answer:
this is not exact because i Estimated the square root of 33
Step-by-step explanation: